What is the difference between a data warehouse and a database? Please two major differences clearly. What are the similarities between a data warehouse and a database? Please two key similarities clearly. (1.5 Marks)

Answer:

hello your question lacks some information attached below is the complete question with the required information

answer : 81.63 mm

Explanation:

settlement of the surface due to compression of the clay ( new consolidated )

= 81.63 mm

attached below is a detailed solution to the given problem

Answer:

The storage of the lake has increased in \(4.58\times 10^{6}\) cubic feet during the month.

Explanation:

We must estimate the monthly storage change of the lake by considering inflows, outflows, seepage and evaporation losses and precipitation. That is:

\(\Delta V_{storage} = V_{inflow} -V_{outflow}-V_{seepage}-V_{evaporation}+V_{precipitation}\)

Where \(\Delta V_{storage}\) is the monthly storage change of the lake, measured in cubic feet.

Monthly inflow

\(V_{inflow} = \left(30\,\frac{ft^{3}}{s} \right)\cdot \left(3600\,\frac{s}{h} \right)\cdot \left(24\,\frac{h}{day} \right)\cdot (30\,days)\)

\(V_{inflow} = 77.76\times 10^{6}\,ft^{3}\)

Monthly outflow

\(V_{outflow} = \left(27\,\frac{ft^{3}}{s} \right)\cdot \left(3600\,\frac{s}{h} \right)\cdot \left(24\,\frac{h}{day} \right)\cdot (30\,days)\)

\(V_{outflow} = 66.98\times 10^{6}\,ft^{3}\)

Seepage losses

\(V_{seepage} = s_{seepage}\cdot A_{lake}\)

Where:

\(s_{seepage}\) - Seepage length loss, measured in feet.

\(A_{lake}\) - Surface area of the lake, measured in square feet.

If we know that \(s_{seepage} = 1.5\,in\) and \(A_{lake} = 525\,acres\), then:

\(V_{seepage} = (1.5\,in)\cdot \left(\frac{1}{12}\,\frac{ft}{in} \right)\cdot (525\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)\)

\(V_{seepage} = 2.86\times 10^{6}\,ft^{3}\)

Evaporation losses

\(V_{evaporation} = s_{evaporation}\cdot A_{lake}\)

Where:

\(s_{evaporation}\) - Evaporation length loss, measured in feet.

\(A_{lake}\) - Surface area of the lake, measured in square feet.

If we know that \(s_{evaporation} = 6\,in\) and \(A_{lake} = 525\,acres\), then:

\(V_{evaporation} = (6\,in)\cdot \left(\frac{1}{12}\,\frac{ft}{in} \right)\cdot (525\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)\)

\(V_{evaporation} = 11.44\times 10^{6}\,ft^{3}\)

Precipitation

\(V_{precipitation} = s_{precipitation}\cdot A_{lake}\)

Where:

\(s_{precipitation}\) - Precipitation length gain, measured in feet.

\(A_{lake}\) - Surface area of the lake, measured in square feet.

If we know that \(s_{precipitation} = 4.25\,in\) and \(A_{lake} = 525\,acres\), then:

\(V_{precipitation} = (4.25\,in)\cdot \left(\frac{1}{12}\,\frac{ft}{in} \right)\cdot (525\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)\)

\(V_{precipitation} = 8.10\times 10^{6}\,ft^{3}\)

Finally, we estimate the storage change of the lake during the month:

\(\Delta V_{storage} = 77.76\times 10^{6}\,ft^{3}-66.98\times 10^{6}\,ft^{3}-2.86\times 10^{6}\,ft^{3}-11.44\times 10^{6}\,ft^{3}+8.10\times 10^{6}\,ft^{3}\)

\(\Delta V_{storage} = 4.58\times 10^{6}\,ft^{3}\)

The storage of the lake has increased in \(4.58\times 10^{6}\) cubic feet during the month.

The volume of water gained and the loss of water through flow,

seepage, precipitation and evaporation gives the storage change.

Response:

Given parameters:

Area of the lake = 525 acres

Inflow = 30 ft.³/s

Outflow = 27 ft.³/s

Seepage loss = 1.5 in. = 0.125 ft.

Total precipitation = 4.25 inches

Evaporator loss = 6 inches

Number of seconds in a month is found as follows;

\(30 \ days/month \times \dfrac{24 \ hours }{day} \times \dfrac{60 \, minutes}{Hour} \times \dfrac{60 \, seconds}{Minute} = 2592000 \, seconds\)

Number of seconds in a month = 2592000 s.

Volume change due to flow, \(V_{fl}\) = (30 ft.³/s - 27 ft.³/s) × 2592000 s = 7776000 ft.³

1 acre = 43560 ft.²

Therefore;

525 acres = 525 × 43560 ft.² =  2.2869 × 10⁷ ft.²

Volume of water in seepage loss, \(V_s\) = 0.125 ft. × 2.2869 × 10⁷ ft.² = 2,858,625 ft.³

Volume gained due to precipitation, \(V_p\) = 0.354167 ft. × 2.2869 × 10⁷ ft.² = 8,099,445.123 ft.³

Volume evaporation loss, \(V_e\) = 0.5 ft. × 2.2869 × 10⁷ ft.² = 11,434,500 ft.³

Which gives;

ΔV = 7776000 - 2858625 + 8099445.123 - 11434500 = 1582823.123

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Answer:

Explanation:

Do you mean food service? It's kind of iffy as an answer. It might be handy to understand the chemistry of foods and the technology of refrigeration. That makes food services hard to classify.

Education: no. You can avoid the math and data skills. You don't need them in education. I knew many teachers that had no idea what a computer did.

technology: definitely. There aren't many technologies that make data processing and mathematics unnecessary.

engineering: definitely.

Answer

Engineering

Technology

maybe food services.

The first two for sure.

To perform the tasks mentioned in the lab, follow the steps below:

On CorpDC:

Log in to the CorpDC server.

Open the Active Directory Users and Computers management console.

Navigate to the Engineering OU.

Create a new group with the following specifications:

Group name: Lead-Engineers

Group members: Emma Tomco and Treg Reese

On CorpServer:

Log in to the CorpServer.

Open the Hyper-V Manager.

Locate the CorpDC2 virtual machine.

Move the storage for the CorpDC2 virtual machine to the H:\HYPERV2 location.

On CorpCluster1:

Log in to the CorpCluster1 server.

Open the Server Manager.

Add the File Server role to the CorpCluster1 server.

Specify the appropriate file server type for virtual machines with extended file open periods.

Name the clustered role as CorpApps.

On CorpCluster1 and CorpCluster2:

Log in to both CorpCluster1 and CorpCluster2 servers.

Open the Failover Cluster Manager.

Configure the cluster quorum settings for CorpCluster using the following specifications:

Add a file share witness.

Use the file share path: \CorpServer\Witness.

Note: Ensure that CorpCluster1 and CorpCluster2 servers are physically located in the Networking Closet of Building A.

By following these steps, you will complete the specified tasks in the lab.

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Create a group named "Lead-Engineers" in the Engineering OU of the BizCorp domain on CorpDC, with members Emma Tomco and Treg Reese. Move CorpDC2's storage to H:\HYPERV2 on CorpServer. Add the File Server role to CorpCluster1, naming it "CorpApps." Configure CorpCluster's quorum settings on CorpCluster1 with a file share witness at I\CorpServer\Witness.corpCluster1.

To perform the tasks outlined in the lab, follow these steps:

1. On CorpDC, create a group within the Engineering OU of the BizCorp domain. Name the group "Lead-Engineers" and add the members Emma Tomco and Treg Reese.

2. On CorpServer, move the storage for the CorpDC2 virtual machine to the H:\HYPERV2 location.

3. On CorpCluster1, add the File Server role to the CorpCluster. Both CorpCluster1 and CorpCluster2 servers are part of the CorpCluster cluster. Name the clustered role "CorpApps."

4. On CorpCluster1, configure the cluster quorum settings for CorpCluster using the following specifications:

  - Add a file share witness.

  - Set the file share path as I\CorpServer\Witness.corpCluster1.

Note: CorpCluster1 and CorpCluster2 are located in the Networking Closet of Building A.

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Answer:

yrt a

Explanation:

Using the knowledge in computer language in python it is possible to write a code that relates the hours of paid work of an employee

Overtime pay

Input pay, rate

If rate =="Hourly" then

 print "Eligible for overtime"

Else if rate =="Monthly" then

 print "Not eligible for overtime"

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The aluminum case of a power transistor that is attached to a square aluminum plate of thermal conductivity k is considered in this problem. A power transistor is a type of transistor that is designed to handle high power levels,

Making it suitable for use in power amplifiers, voltage regulators, and other applications that require high currents and voltages.The thermal conductivity of a material is a measure of how well it conducts heat. The higher the thermal conductivity of a material, the better it is at transferring heat from one place to another. Aluminum is a good conductor of heat, with a thermal conductivity of around 200 W/m·K.

This means that it can transfer heat quickly and efficiently from the transistor to the plate.The problem asks us to consider the heat transfer from the transistor to the plate, and to determine the temperature rise of the transistor as a result.

We can use the following equation to calculate the temperature rise:

ΔT = P * Rθ

where ΔT is the temperature rise in degrees Celsius, P is the power dissipated by the transistor in watts, and Rθ is the thermal resistance from the transistor to the plate in degrees Celsius per watt.

The thermal resistance can be calculated using the following equation:

Rθ = t/kA

where t is the thickness of the aluminum case, k is the thermal conductivity of aluminum, and A is the surface area of the aluminum case. We can assume that the thickness of the case is uniform and that the surface area is equal to the area of the base of the case, since this is where the case is attached to the plate.

The power dissipated by the transistor can be calculated using the following equation:

P = VCE * IC

where VCE is the voltage across the collector and emitter terminals of the transistor, and IC is the current flowing through the collector terminal. We can assume that the voltage and current are constant, since the transistor is designed to operate at a specific voltage and current level.

To calculate the temperature rise of the transistor, we need to determine the values of P and Rθ, and then substitute them into the equation for ΔT. We can then solve for ΔT to obtain the temperature rise in degrees Celsius.

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Answer:

  5

Explanation:

One pulse is required for each bit to be moved.

5 pulses are required.

Dimensioning standards for inches and millimeters are used to specify the size and location of features on an object or part. The primary difference between these two standards is the unit of measurement used.

Inches are the primary unit of measurement in the United States, and dimensioning standards for inches are based on the imperial system of measurement. This system is based on units of inches, feet, and yards.

Dimensioning standards for inches typically use fractions of an inch, such as 1/8", 1/16", or 1/32", to specify dimensions. These fractions are commonly used because they are easy to measure with common tools like rulers and calipers.

On the other hand, millimeters are the primary unit of measurement in most other parts of the world, and dimensioning standards for millimeters are based on the metric system of measurement. This system is based on units of millimeters, centimeters, and meters.

Dimensioning standards for millimeters typically use decimals, such as 1.5 mm or 3.75 mm, to specify dimensions. Decimals are commonly used in the metric system because they allow for more precise measurements and are easier to work with in mathematical calculations.

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Answer:

frejfru8758u37

Explanation:

i need points

The normal stress on the inclined plane ab is σ = 2.0625 ksi and the shear stress on the inclined plane ab is τ = 1.375 ksi.

Determining the normal stress and shear stress acting on the inclined plane ab, use the stress transformation equations.

The stress transformation equations relate the normal stress, σ, and shear stress, τ, on an inclined plane with the corresponding normal and shear stresses acting on a plane perpendicular to the inclined plane. The equations are:

σ = σn cos²θ + σs sin²θ ± 2τnsinθcosθ

τ = (σn - σs) sinθcosθ ± τn cos²θ ± τs sin²θ

where θ is the angle between the inclined plane and the plane of reference, σn and σs are the normal stresses acting on the plane of reference, τn and τs are the shear stresses acting on the plane of reference, and τns is the shear stress acting on the inclined plane.

Given that τ = 5.5 ksi and σ and τ are to be determined on the inclined plane ab. Assume a plane of reference perpendicular to the inclined plane ab with angle θ in between.

The angle θ between the inclined plane and the plane of reference is the complement of the angle of inclination, so θ = 90° - 30° = 60°.

Assuming that there are no normal stresses acting on the plane of reference, so σn = 0. Therefore, the stress transformation equations simplify to:

σ = σs sin²θ ± 2τnsinθcosθ

τ = (σs/2) ± τn cos²θ ± (τ/2)sin²θ

Solve for σs and τn by using the given value of τ and the fact that there are no normal stresses acting on the plane of reference:

τ = τns = 5.5 ksi

σs = τ/2 = 2.75 ksi

τn = 0

Substitute these values into the stress transformation equations to obtain the normal stress and shear stress on the inclined plane ab:

σ = σs sin²θ = 2.75 ksi sin²60° = 2.75 ksi (0.75) = 2.0625 ksi

τ = (σs/2) ± τn cos²θ ± (τ/2)sin²θ = 2.75 ksi/2 = 1.375 ksi

Therefore, the normal stress on the inclined plane ab is σ = 2.0625 ksi and the shear stress on the inclined plane ab is τ = 1.375 ksi.

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Langasite (LGS) has become a subject of interest for its applications in the area of the development of high-temperature sensors for harsh environments. The high-temperature static strain Langasite (LGS) sawr sensor is one such development.
Temperature Compensation: Langasite (LGS) is a piezoelectric material that produces a voltage when subjected to a mechanical strain. When exposed to high temperatures, Langasite (LGS) sensors may experience errors due to the changes in the physical properties of the material.

Numerical Calibration for Direct Strain Reading:Langasite (LGS) sensors must be calibrated to obtain accurate readings. Numerical calibration is a technique that is commonly used to calibrate Langasite (LGS) sawr sensors.
In conclusion, the high-temperature static strain Langasite (LGS) sawr sensor is an important development that has various applications in the area of high-temperature sensing.

Temperature compensation is required to minimize errors due to changes in the physical properties of the material, while numerical calibration is used to convert raw data into strain values. The use of LGS sensors in harsh environments has the potential to improve the accuracy and reliability of high-temperature sensing.

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Answer:

Hence, the steady state temperature distribution between concentric spheres with radii 1 and 4 is:

u = \(\frac{C1}{r }\) + C2

u = \(\frac{\frac{-320}{3} }{r}\) + \(\frac{320}{3}\)

Explanation:

Solution:

Note: This question is incomplete and lacks necessary data to solve this question. But I have found a similar question and will try to solve it.

This question has this following part missing:

"If the temperature of the outer sphere is maintained at 80 degrees and the inner sphere at 0 degrees.

Now, this question is complete.

Let's find out the steady state temperature distribution.

As, we know the spherical symmetric heat equation:

\(\frac{du}{dt}\) = \(\frac{k}{r^{2} }\) \(\frac{D}{Dr}\) (\(r^{2}\) \(\frac{Du}{Dr}\))

Where, small (d) represents the partial differentiation and Capital D represents simple derivative.

And the ordinary differential equation (ODE) for steady state temperature distribution is:

\(\frac{k}{r^{2} }\) \(\frac{D}{Dr}\)(\(r^{2}\) \(\frac{Du}{Dr}\)) = 0

So it can be said that:

\(r^{2}\) \(\frac{Du}{Dr}\)

Consequently,

\(\frac{Du}{Dr}\) = \(\frac{C1}{r^{2} }\)

Taking derivative of the above equation we get:

u = -\(\frac{C1}{r }\) + C2

Solution of the ordinary differential equation

u = \(\frac{C1}{r }\) + C2 (Consuming the negative sign into C1 constant)

As, it is given in our question that our boundary conditions are: 1 and 4

So,

Putting the boundary conditions into the variable (r) to find the constants we get:

u1 = \(\frac{C1}{1} }\) + C2 = 0 (degrees)

u1 = C1 + C2 = 0 (degrees)  equation (1)

Similarly,

u4 = \(\frac{C1}{4}\) + C2 = 80 (degrees)

u4 = \(\frac{C1}{4}\) + C2 = 80 (degrees)  equation (2)

Solving for C1, we get:

Equation 1 - Equation 2

(C1 - C2 = 0) - (\(\frac{C1}{4}\) + C2 = 80)

\(\frac{-3}{4}\)C1 = 80

Solving for C1

C1 = -\(\frac{320}{3}\)

With the help of value of C1, we get value of C2

Put the value of C1 in equation (1) to get value of C2

C1 + C2 = 0

-\(\frac{320}{3}\) + C2 = 0

Solving for C2

C2 = \(\frac{320}{3}\)

Hence, the steady state temperature distribution between concentric spheres with radii 1 and 4 is:

u = \(\frac{C1}{r }\) + C2

Plugging in the values of C1 and C2

u = \(\frac{\frac{-320}{3} }{r}\) + \(\frac{320}{3}\)

Inadequate nutrition, pollution-associated health disorders and communicable illnesses, poor sanitation and housing conditions, and linked health conditions are some of the primary health concerns caused by urbanization. Hence, a properly planned topography will prevent these negative concerns.

The drafting of plans for, as well as the control and management of, towns, cities, and metropolitan regions falls within the purview of urban planning. It tries to arrange sociospatial linkages at many levels of government and governance.

Urban planning can help to create more livable and desirable communities by designing for walkability, accessibility, and green spaces.

It can help to create more efficient and sustainable communities by promoting the use of public transportation, promoting mixed-use development, and designing for energy efficiency. This can help to reduce greenhouse gas emissions, reduce traffic congestion, and save energy.

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Technical practice incorporates build-time identification of security vulnerabilities in the code is  Penetration testing.

A penetration test, sometimes referred to as a pen test or ethical hacking, is a legitimate simulated cyberattack on a computer system that is carried out to analyze the system's security. This is distinct from a vulnerability assessment.

In order to identify and illustrate the financial effects of a system's vulnerabilities, penetration testers employ the same tools, strategies, and procedures as attackers. Reconnaissance, scanning, vulnerability assessment, exploitation, and reporting are the five stages of a penetration test.

Penetration testing is a technical activity that includes build-time discovery of security vulnerabilities in the code.

Penetration tests are essential to an organization's security because they teach staff members how to respond to any kind of intrusion from a malicious party. Pen tests are a method of determining whether a company's security procedures are actually effective.

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The flowrate Q is estimated to be 10.83 m³/s, and the uncertainty is 0.64 m³/s (i.e., 5.89%).

To estimate the uncertainty in the flowrate for the measured values: Q = 128μL ΔPπD / 4.

Where ΔP = 12kN/m² ± 0.3kN/m²; D = 0.3m ± 0.01 m,

L = 150m ± 0.1m and μ = 0.001 kg/m-s (the dynamic viscosity of water).

The formula for the uncertainty of flowrate using the given parameters can be calculated as follows:

ΔQ/Q = ΔΔP/ΔP + ΔD/D + ΔL/L Since ΔP = 12kN/m² ± 0.3kN/m²,

The percentage uncertainty in ΔP is given as follows: (0.3kN/m²/12kN/m²) × 100% = 2.5%

Similarly, the percentage uncertainty for D and L can be computed as follows:

ΔD/D = (0.01m/0.3m) × 100% = 3.33%ΔL/L = (0.1m/150m) × 100% = 0.067%

The flowrate Q can be calculated as follows: Q = 128μL ΔPπD / 4Q = 128 x 0.001 x 150 x (12/4) x 3.1416 x (0.3) ⁴ = 10.83 m³/s

The uncertainty in Q can be calculated as follows:

ΔQ/Q = ΔΔP/ΔP + ΔD/D + ΔL/LΔQ/Q = 2.5% + 3.33% + 0.067%ΔQ/Q = 5.89%

Therefore, the flowrate uncertainty is ΔQ = (5.89/100) x 10.83 = 0.64 m³/s.

The volumetric flowrate formula in a horizontal pipe for laminar flow is Q = 128 μL ΔPπD / 4.

The flowrate uncertainty can be calculated using the formula, ΔQ/Q = ΔΔP/ΔP + ΔD/D + ΔL/L.

The values of ΔP, D, and L are ΔP = 12k N/m ² ± 0.3k N/m ², D = 0.3m ± 0.01 m, and L = 150m ± 0.1m,

Therefore, the flowrate Q is estimated to be 10.83 m³/s, and the uncertainty is 0.64 m³/s (i.e., 5.89%).

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Answer:

Explanation:

Civil engineers have become experts in creating sustainable and environmentally friendly buildings and systems. Multiplied over many communities, the energy and emissions savings can make a real difference in the environment. Other life-improving functions can also make communities better places to live.Jul 19, 2017

To determine the circulation for the given vector field around the enclosed half circle of radius 1, we need to calculate the line integral of the vector field along the boundary of the half circle.

The line integral is calculated by integrating the dot product of the vector field and the differential of the boundary. The circulation is then equal to the line integral divided by 2π. To calculate the line integral, we need to know the vector field and the boundary of the half circle. Once we have this information, we can use the equation for the line integral to calculate the circulation.

To calculate the circulation for the given vector field around the enclosed half circle of radius 1, we need to first calculate the line integral of the vector field along the boundary of the half circle. The line integral is calculated by integrating the dot product of the vector field and the differential of the boundary. The differential of the boundary is equal to the vector from the starting point to the end point of the boundary. The vector field can be expressed as a function of the coordinates of the boundary. Once we have the vector field and the differential of the boundary, we can use the equation for the line integral to calculate the circulation. The circulation is then equal to the line integral divided by 2π.

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Answer:

Space velocity = 30 hr⁻¹

Explanation:

Space velocity for reactors express how much reactor volume of feed or reactants can be treated per unit time. For example, a space velocity of 3 hr⁻¹ means the reactor can process 3 times its volume per hour.

It is given mathematically as

Space velocity = (volumetric flow rate of the reactants)/(the reactor volume)

Volumetric flowrate of the reeactants

= (molar flow rate)/(concentration)

Molar flowrate of the reactants = 300 millimol/hr

Concentration of the reactants = 100 millimol/liter

Volumetric flowrate of the reactants = (300/100) = 3 liters/hr

Reactor volume = 0.1 liter

Space velocity = (3/0.1) = 30 /hr = 30 hr⁻¹

Hope this Helps!!!

Answer:

it should be A bc I took the test a while back

When the power rating of a transformer is exceeded by placing too great on the transformer, the voltage will start to drop, and the transformer may overheat and be damaged.

This is because exceeding the power rating causes the transformer to draw more current than it is designed to handle, resulting in increased losses due to the resistance of the winding and core materials. This increased current also causes a voltage drop across the winding and leads to a decrease in output voltage, which can cause problems downstream in the circuit. If the current continues to increase beyond the transformer's capacity, it can cause overheating and damage to the winding insulation, leading to short circuits and potential safety hazards.

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Answer: B, plumb bob

Explanation:

A plumb bob's main job was to establish what is true (ie vertical and perpendicular to a surface,) as well as the secondary task of marking these points.

Compasses are mainly used to draw circles, protractors are used to measure angles, and rulers are used to measuring length.

Answer:

Some good ways and popular ways are:

1. Investing in data literacy.

2. Think about cybersecurity.

3. Choosing good and the right tools.

4. Establish metrics that matter.

Explanation:

These all are helpful.

(Hope this helped! Have a great day.)

Answer:

Hello your question is incomplete attached below is the complete question

Answer : Factor of safety for point A :

i) using MSS

(Fos)MSS =  3.22

ii) using DE

(Fos)DE = 3.27

Factor of safety for point B

i) using MSS

(Fos)MSS =  3.04

ii) using DE

(Fos)DE = 3.604

Explanation:

Factor of safety for point A :

i) using MSS

(Fos)MSS =  3.22

ii) using DE

(Fos)DE = 3.27

Factor of safety for point B

i) using MSS

(Fos)MSS =  3.04

ii) using DE

(Fos)DE = 3.604

Attached below is the detailed solution

Answer:

545454cm

Explanation:

The blue print drawing of the house shows 666 centimeters, but the real picture of the house is 333 meters. So let the number of cm on the blueprint that represent the distance of 272727 meters be x. Firstly convert the meters to centimeters 666cm = 333m, x=272727m ; then cross multiply, 666cm=33300cm x=27272700cm ; x =(666cm×272727cm)/33300cm =545454cm.

Answer:

control loop unit

Explanation:

Edmentum/Plato

Answer:

Adjective. time-dependent (not comparable) (mathematics, physics) Determined by the value of a variable representing time.

Explanation:

Plz mark brainliest thanks

Answer: Adjective. time-dependent (not comparable) (mathematics, physics) Determined by the value of a variable representing time

Answer: Kirchhoff's voltage law